The determinant of (17) at the equilibrium point is given by
det{J}_{T}(\overline{x},\overline{y})=\frac{4{b}_{1}\overline{y}({A}_{1}{c}_{2}{a}_{2})}{\overline{x}{({\overline{y}}^{2}+{A}_{1})}^{2}}.
(18)
The trace of (17) at the equilibrium point is given by
tr{J}_{T}(\overline{x},\overline{y})=\frac{2{c}_{2}\overline{y}}{{\overline{x}}^{2}}+2.
The characteristic equation has the form
{\lambda}^{2}\lambda (\frac{2{c}_{2}\overline{y}}{{\overline{x}}^{2}}+2)+\frac{4{b}_{1}\overline{y}({A}_{1}{c}_{2}{a}_{2})}{\overline{x}{({\overline{y}}^{2}+{A}_{1})}^{2}}=0.
Equilibrium curves {\mathcal{C}}_{f}=\{(x,y)\in \mathcal{R}:f(x,y)=x\} and {\mathcal{C}}_{g}=\{(x,y)\in \mathcal{R}:g(x,y)=y\} can be given explicitly as functions of y:
\begin{array}{c}{\mathcal{C}}_{f}:{x}_{f}(y)=\frac{{A}_{1}+{y}^{2}}{{b}_{1}},\hfill \\ {\mathcal{C}}_{g}:\{\begin{array}{ll}{x}_{{g}^{+}}(y)=+\sqrt{\frac{{a}_{2}+{c}_{2}{y}^{2}}{y}},& y>0,\\ {x}_{{g}^{}}(y)=\sqrt{\frac{{a}_{2}+{c}_{2}{y}^{2}}{y}},& y>0.\end{array}\hfill \end{array}
Note that {x}_{{g}^{}}(x) is always negative. Let {x}_{g}(y) denote {x}_{{g}^{+}}(y). We consider only {x}_{f}(y) and {x}_{g}(y). Let
\tilde{x}(y)={x}_{f}(y){x}_{g}(y).
Lemma 3 Let T=(f,g) be the map defined by (5). Then {f}_{x}^{\prime}(x,y)>1, and the following is true:
sign(\tilde{x}(y))=sign({y}^{5}+2{A}_{1}{y}^{3}{b}_{1}^{2}{c}_{2}{y}^{2}+{A}_{1}^{2}y{a}_{2}{b}_{1}^{2}).
Proof The first derivative of {x}_{f}(y) is given by
{x}_{f}^{\prime}(y)=\frac{{f}_{y}^{\prime}(x,y)}{1{f}_{x}^{\prime}(x,y)}=\frac{2y}{{b}_{1}}>0.
Since {f}_{y}^{\prime}(x,y)<0, we get {f}_{x}^{\prime}(x,y)>1. Further,
\tilde{x}(y)={x}_{f}(y){x}_{g}(y)=\frac{{A}_{1}+{y}^{2}}{{b}_{1}}\sqrt{\frac{{a}_{2}+{c}_{2}{y}^{2}}{y}}=\frac{\sqrt{y}({A}_{1}+{y}^{2}){b}_{1}\sqrt{{a}_{2}+{c}_{2}{y}^{2}}}{{b}_{1}\sqrt{y}}.
Now, the proof follows from
{\left(\sqrt{y}({A}_{1}+{y}^{2})\right)}^{2}{\left({b}_{1}\sqrt{{a}_{2}+{c}_{2}{y}^{2}}\right)}^{2}={y}^{5}+2{A}_{1}{y}^{3}{b}_{1}^{2}{c}_{2}{y}^{2}+{A}_{1}^{2}y{a}_{2}{b}_{1}^{2}.
□
Lemma 4 Let T be the map defined by (5), and let
{J}_{T}(\overline{x},\overline{y})=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)
(19)
be the Jacobian matrix of T at a fixed point (\overline{x},\overline{y}). Then the Jacobian matrix (19) has real and distinct eigenvalues {\lambda}_{1} and {\lambda}_{2} such that {\lambda}_{1}<{\lambda}_{2} and {\lambda}_{2}>1. Furthermore, the following holds:
sign({\tilde{x}}^{\prime}(\overline{y}))=sign(1{\lambda}_{1}).
Proof Implicit differentiation of the equations defining {C}_{f} and {C}_{g} at (\overline{x},\overline{y}) gives
{x}_{f}^{\prime}(\overline{y})=\frac{{f}_{y}^{\prime}(\overline{x},\overline{y})}{1{f}_{x}^{\prime}(\overline{x},\overline{y})},\phantom{\rule{2em}{0ex}}{x}_{g}^{\prime}(\overline{y})=\frac{1{g}_{y}^{\prime}(\overline{x},\overline{y})}{{g}_{x}^{\prime}(\overline{x},\overline{y})}.
(20)
The characteristic equation associated with the Jacobian matrix of T at (\overline{x},\overline{y}) is given by
\begin{array}{rl}p(\lambda )& ={\lambda}^{2}[{f}_{x}^{\prime}(\overline{x},\overline{y})+{g}_{y}^{\prime}(\overline{x},\overline{y})]\lambda +[{f}_{x}^{\prime}(\overline{x},\overline{y}){g}_{y}^{\prime}(\overline{x},\overline{y}){f}_{y}^{\prime}(\overline{x},\overline{y}){g}_{x}^{\prime}(\overline{x},\overline{y})]\\ ={\lambda}^{2}(a+d)\lambda +(adbc).\end{array}
Since the map T is competitive, then the eigenvalues of the Jacobian matrix of the map T at the equilibrium (\overline{x},\overline{y}) are real and distinct and {\lambda}_{1}<{\lambda}_{2}. By (20), we have
\begin{array}{rl}{\tilde{x}}^{\prime}(\overline{y})& ={x}_{f}^{\prime}(\overline{y}){x}_{g}^{\prime}(\overline{y})=\frac{{f}_{y}^{\prime}(\overline{x},\overline{y})}{1{f}_{x}^{\prime}(\overline{x},\overline{y})}\frac{1{g}_{y}^{\prime}(\overline{x},\overline{y})}{{g}_{x}^{\prime}(\overline{x},\overline{y})}\\ =\frac{b}{1a}\frac{1d}{c}=\frac{1+(a+d)(adbc)}{c(1a)}\\ =\frac{p(1)}{c(1a)}=\frac{(1{\lambda}_{1})(1{\lambda}_{2})}{c(a1)}.\end{array}
From tr{J}_{T}(\overline{x},\overline{y})={\lambda}_{1}+{\lambda}_{2}>2 we get {\lambda}_{2}>1. The map T is competitive, which implies c={g}_{x}^{\prime}(\overline{x},\overline{y})<0. In view of Lemma 3, we get a={f}_{x}^{\prime}(\overline{x},\overline{y})>1, from which it follows that sign({\tilde{x}}^{\prime}(\overline{y}))=sign(1{\lambda}_{1}). □
Theorem 10 Assume that {\mathrm{\Delta}}_{1}>0 and {\lambda}_{1} and {\lambda}_{2} are eigenvalues of {J}_{T}(\overline{x},\overline{y}). Then there exists the unique equilibrium point E=(\overline{x},\overline{y}) and the following hold:

(a)
If {a}_{2}<{A}_{1}{c}_{2}, then E is a saddle point and 0<{\lambda}_{1}<1, {\lambda}_{2}>1.

(b)
Assume that {a}_{2}>{A}_{1}{c}_{2}. Let
\mathrm{\Gamma}(\overline{y})=5{c}_{2}{\overline{y}}^{4}+{\overline{y}}^{2}(9{A}_{1}{c}_{2}{a}_{2})+3{a}_{2}{A}_{1}.

(b1)
If \mathrm{\Gamma}(\overline{y})>0, then E is a saddle point. Furthermore, the following hold: 1<{\lambda}_{1}<0, {\lambda}_{2}>1.

(b2)
If \mathrm{\Gamma}(\overline{y})<0, then E is a repeller. Furthermore, the following hold: {\lambda}_{1}<1, {\lambda}_{2}>1; {\lambda}_{1}<{\lambda}_{2}.

(b3)
If \mathrm{\Gamma}(\overline{y})=0, then E is a nonhyperbolic equilibrium point. Furthermore, the following hold: {\lambda}_{1}=1, {\lambda}_{2}>1.
Proof In view of (7) and Lemma 1, we have that the function
\tilde{f}(y)={y}^{5}+2{A}_{1}{y}^{3}{b}_{1}^{2}{c}_{2}{y}^{2}+{A}_{1}^{2}y{a}_{2}{b}_{1}^{2}
has one zero \overline{y} of multiplicity one. In view of Lemma 1, the map T has a unique equilibrium point. Since \tilde{f}(0)={a}_{2}{b}_{1}^{2}<0 and {lim}_{y\to +\mathrm{\infty}}\tilde{f}(y)=+\mathrm{\infty}, we have \tilde{f}(y)<0 for y<\overline{y} and \tilde{f}(y)>0 for y>\overline{y}. By Lemmas 6 and 7 from [27], the equilibrium curves {\mathcal{C}}_{f} and {\mathcal{C}}_{g} intersect transversally at (\overline{x},\overline{y}), i.e., {\tilde{x}}^{\prime}(\overline{y})\ne 0. In view of Lemma 3 and by the continuity of function \tilde{x}(y), there exists a neighborhood {U}_{\overline{y}} of \overline{y} such that {\tilde{x}}^{\prime}(y)>0 for y\in {U}_{\overline{y}}, which implies
{\tilde{x}}^{\prime}(\overline{y})={x}_{f}^{\prime}(\overline{y}){x}_{g}^{\prime}(\overline{y})>0.
(21)
From (21) and Lemma 4 we obtain {\lambda}_{1}<1 and {\lambda}_{2}>1.
If {a}_{2}<{A}_{1}{c}_{2}, then det{J}_{T}(\overline{x},\overline{y})={\lambda}_{1}{\lambda}_{2}>0, which implies that {\lambda}_{1}\in (0,1).
Now, assume that {a}_{2}>{A}_{1}{c}_{2}. By using
{b}_{1}=\frac{{\overline{y}}^{2}+{A}_{1}}{\overline{x}},\phantom{\rule{2em}{0ex}}\overline{x}=\sqrt{\frac{{c}_{2}{\overline{y}}^{2}+{a}_{2}}{\overline{y}}},
one can see that
\begin{array}{rl}p(1)& =1+det{J}_{T}(\overline{x},\overline{y})+tr{J}_{T}(\overline{x},\overline{y})=\frac{5{c}_{2}{\overline{y}}^{4}+{\overline{y}}^{2}(9{A}_{1}{c}_{2}{a}_{2})+3{a}_{2}{A}_{1}}{({\overline{y}}^{2}+{A}_{1})({c}_{2}{\overline{y}}^{2}+{a}_{2})}\\ =\frac{\mathrm{\Gamma}(\overline{y})}{({\overline{y}}^{2}+{A}_{1})({c}_{2}{\overline{y}}^{2}+{a}_{2})}\end{array}
and
det{J}_{T}(\overline{x},\overline{y})={\lambda}_{1}{\lambda}_{2}=\frac{4{b}_{1}\overline{y}({A}_{1}{c}_{2}{a}_{2})}{\overline{x}{({\overline{y}}^{2}+{A}_{1})}^{2}},
where p(\lambda )=(\lambda {\lambda}_{1})(\lambda {\lambda}_{2}). In view of (18) and p(1)=({\lambda}_{1}+1)({\lambda}_{2}+1), we obtain statement (b) of the theorem. □
Lemma 5 Suppose that all the assumptions of Theorem 10 are satisfied. Let
{y}_{\pm}=\sqrt{\frac{{a}_{2}9{A}_{1}{c}_{2}\pm \sqrt{78{a}_{2}{A}_{1}{c}_{2}+{a}_{2}^{2}+81{A}_{1}^{2}{c}_{2}^{2}}}{10{c}_{2}}}.
Then the following statements are true.

(a)
\mathrm{\Gamma}(\overline{y})>0 if and only if one of the following inequalities holds:
\begin{array}{c}9{A}_{1}{c}_{2}{a}_{2}\ge 0,\hfill \\ 9{A}_{1}{c}_{2}{a}_{2}<0\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}78{a}_{2}{A}_{1}{c}_{2}+{a}_{2}^{2}+81{A}_{1}^{2}{c}_{2}^{2}<0,\hfill \\ 9{A}_{1}{c}_{2}{a}_{2}<0\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}78{a}_{2}{A}_{1}{c}_{2}+{a}_{2}^{2}+81{A}_{1}^{2}{c}_{2}^{2}\ge 0\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\hfill \\ (\tilde{f}({y}_{})>0\mathit{\text{or}}\tilde{f}({y}_{+})0);\hfill \end{array}

(b)
\mathrm{\Gamma}(\overline{y})<0 if and only if the following hold:
9{A}_{1}{c}_{2}{a}_{2}<0\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}78{a}_{2}{A}_{1}{c}_{2}+{a}_{2}^{2}+81{A}_{1}^{2}{c}_{2}^{2}\ge 0\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}(\tilde{f}({y}_{})<0\mathit{\text{and}}\tilde{f}({y}_{+})0);

(c)
\mathrm{\Gamma}(\overline{y})=0
if and only if
9{A}_{1}{c}_{2}{a}_{2}<0\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}78{a}_{2}{A}_{1}{c}_{2}+{a}_{2}^{2}+81{A}_{1}^{2}{c}_{2}^{2}\ge 0\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}(\tilde{f}({y}_{})=0\mathit{\text{or}}\tilde{f}({y}_{+})=0).
Proof The function \tilde{f}(y) has one simple zero \overline{y}, which implies \tilde{f}(y)<0 for 0\le y<\overline{y} and \tilde{f}(y)>0 for y>\overline{y}. Then
\overline{y}>\alpha \phantom{\rule{1em}{0ex}}\text{if and only if}\phantom{\rule{1em}{0ex}}\tilde{f}(\alpha )<0,
while
\overline{y}<\beta \phantom{\rule{1em}{0ex}}\text{if and only if}\phantom{\rule{1em}{0ex}}\tilde{f}(\beta )>0
for some \alpha ,\beta \in [0,\mathrm{\infty}). Now the proof follows from the fact that \tilde{f}(y)=0 has real roots
\begin{array}{c}\{\sqrt{\frac{{a}_{2}9{A}_{1}{c}_{2}\pm \sqrt{78{a}_{2}{A}_{1}{c}_{2}+{a}_{2}^{2}+81{A}_{1}^{2}{c}_{2}^{2}}}{10{c}_{2}}},\hfill \\ \phantom{\rule{1em}{0ex}}\sqrt{\frac{{a}_{2}9{A}_{1}{c}_{2}\pm \sqrt{78{a}_{2}{A}_{1}{c}_{2}+{a}_{2}^{2}+81{A}_{1}^{2}{c}_{2}^{2}}}{10{c}_{2}}}\}\hfill \end{array}
if and only if
9{A}_{1}{c}_{2}{a}_{2}<0\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}78{a}_{2}{A}_{1}{c}_{2}+{a}_{2}^{2}+81{A}_{1}^{2}{c}_{2}^{2}\ge 0.
□
Theorem 11 Assume that {\mathrm{\Delta}}_{1}<0. Then there exist three distinct equilibrium points in the positive quadrant: {E}_{1}=({\overline{x}}_{1},{\overline{y}}_{1}), {E}_{2}=({\overline{x}}_{2},{\overline{y}}_{2}) and {E}_{3}=({\overline{x}}_{3},{\overline{y}}_{3}) such that {E}_{1}{\ll}_{ne}{E}_{2}{\ll}_{ne}{E}_{3} and the following hold:

(a)
{E}_{1} and {E}_{3} are saddle points. If {\lambda}_{1}^{(i)} and {\lambda}_{2}^{(i)} are the eigenvalues of {J}_{T}({E}_{i}), i=1,3, then 0<{\lambda}_{1}^{(i)}<1, {\lambda}_{2}^{(i)}>1.

(b)
The equilibrium point {E}_{2} is a repeller. If {\lambda}_{1}^{(2)} and {\lambda}_{2}^{(2)} are the eigenvalues of {J}_{T}({E}_{2}), then 1<{\lambda}_{1}^{(2)}<{\lambda}_{2}^{(2)}.
Proof In view of Lemma 1, equation (7) has three positive roots of multiplicity one. Since
\overline{x}=\frac{{A}_{1}+{\overline{y}}^{2}}{{b}_{1}}>0,
then by (a) of Lemma 1 we obtain that the map T has three equilibrium points that we denote by {E}_{1}, {E}_{2} and {E}_{3}. Given points lie on the increasing curve
{x}_{f}(y)=\frac{{A}_{1}+{y}^{2}}{{b}_{1}},
which implies that the points are in the northeast ordering. Descartes’ rule of signs and (8) imply that det{J}_{T}(x,y)>0 when {a}_{2}<{A}_{1}{c}_{2}. In view of (7) and Lemma 1, we have that the polynomial
\tilde{f}(y)={y}^{5}+2{A}_{1}{y}^{3}{b}_{1}^{2}{c}_{2}{y}^{2}+{A}_{1}^{2}y{a}_{2}{b}_{1}^{2}
has three zeros {\overline{y}}_{i}, i=1,2,3, of multiplicity one. Since \tilde{f}(0)={a}_{2}{b}_{1}^{2}<0 and {lim}_{y\to +\mathrm{\infty}}\tilde{f}(y)=+\mathrm{\infty}, we have \tilde{f}(y)<0 for y\in (0,{\overline{y}}_{1})\cup ({\overline{y}}_{2},{\overline{y}}_{3}) and \tilde{f}(y)>0 for y\in ({\overline{y}}_{1},{\overline{y}}_{2})\cup ({\overline{y}}_{3},+\mathrm{\infty}).
By Lemmas 6 and 7 from [27], the equilibrium curves {\mathcal{C}}_{f} and {\mathcal{C}}_{g} intersect transversally at {E}_{1}, {E}_{2} and {E}_{3}, i.e., {\tilde{x}}^{\prime}({\overline{y}}_{i})\ne 0, i=1,2,3. By this and Lemma 3 and by the continuity of function \tilde{x}(y), there exists a neighborhood {U}_{{\overline{y}}_{i}}^{(i)} of {\overline{y}}_{i} such that {\tilde{x}}^{\prime}(y)>0 for y\in {U}_{{\overline{y}}_{i}}^{(i)} for i=1,3 and {\tilde{x}}^{\prime}(y)<0 for y\in {U}_{{\overline{y}}_{i}}^{(2)}. Using this we get
{\tilde{x}}^{\prime}({\overline{y}}_{i})>0\phantom{\rule{1em}{0ex}}\text{for}i=1,3\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{\tilde{x}}^{\prime}({\overline{y}}_{i})0\phantom{\rule{1em}{0ex}}\text{for}i=2.
Let
{J}_{T}({E}_{i})=\left(\begin{array}{cc}{a}_{i}& {b}_{i}\\ {c}_{i}& {d}_{i}\end{array}\right),\phantom{\rule{1em}{0ex}}i=1,2,3.
In view of (18), we have det{J}_{T}({E}_{i})={\lambda}_{1}^{(i)}{\lambda}_{2}^{(i)}>0, i=1,2,3. By Lemma 4 we obtain 0<{\lambda}_{1}^{(i)}<1 and {\lambda}_{2}^{(i)}>1 for i=1,3. Since {\tilde{x}}^{\prime}({\overline{y}}_{2})<0, by Lemma 4 we have 1<{\lambda}_{1}^{(2)}<{\lambda}_{2}^{(2)}. This completes the proof. □
Theorem 12 Assume that {\mathrm{\Delta}}_{1}=0 and {\mathrm{\Delta}}_{2}\ne 0 Then there exist two distinct equilibrium points in the positive quadrant {E}_{1}=({\overline{x}}_{1},{\overline{y}}_{1}) and {E}_{3}=({\overline{x}}_{3},{\overline{y}}_{3}) such that {E}_{1}{\ll}_{ne}{E}_{3}. Let {\lambda}_{1}^{(i)} and {\lambda}_{2}^{(i)} be the eigenvalues of {J}_{T}({E}_{i}), i=1,3. Then the following hold:

(a)
Exactly one of the roots {\overline{y}}_{1} or {\overline{y}}_{3} of (7) has multiplicity two.

(b)
If {\overline{y}}_{1} is a root of (7) of multiplicity two, then the equilibrium point {E}_{1} is nonhyperbolic and {E}_{3} is a saddle point. Furthermore, {\lambda}_{1}^{(1)}=1, {\lambda}_{2}^{(1)}>1 and 0<{\lambda}_{1}^{(3)}<1, {\lambda}_{2}^{(3)}>1.

(c)
If {\overline{y}}_{3} is a root of (7) of multiplicity two, then the equilibrium point {E}_{3} is nonhyperbolic and {E}_{1} is a saddle point. Furthermore, {\lambda}_{1}^{(3)}=1, {\lambda}_{2}^{(3)}>1 and 0<{\lambda}_{1}^{(1)}<1, {\lambda}_{2}^{(1)}>1.
Proof In view of Lemma 1, equation (7) has two positive zeros, one of multiplicity one and another one of multiplicity two, which implies statement (a). Since \overline{x}=({A}_{1}+{\overline{y}}^{2})/{b}_{1}>0, we obtain that the map T has two equilibrium points that we denote by {E}_{1} and {E}_{3}. Descartes’ rule of signs and (8) imply that {a}_{2}<{A}_{1}{c}_{2}\Rightarrow det{J}_{T}(x,y)>0. Now, we prove statement (b). Similarly as in the proof of Theorem 11, one can see that {E}_{3} is a saddle point. In view of Lemmas 6 and 7, from [27] we have that {\tilde{x}}^{\prime}({\overline{y}}_{1})=0, since {\overline{y}}_{1} is the root of (7) of multiplicity two. By Lemma 4 we obtain {\lambda}_{1}^{(1)}=1, {\lambda}_{2}^{(1)}>1. The proof of statement (c) is similar and we will skip it. □
Theorem 13 Assume that {\mathrm{\Delta}}_{1}=0 and {\mathrm{\Delta}}_{2}=0. Then there exists one equilibrium point in the positive quadrant {E}_{1}=({\overline{x}}_{1},{\overline{y}}_{1}) which is nonhyperbolic. If {\lambda}_{1}^{(1)} and {\lambda}_{2}^{(1)} are eigenvalues of {J}_{T}({E}_{1}), then {\lambda}_{1}^{(1)}=1, {\lambda}_{2}^{(1)}>1.
Proof In view of Lemma 1, {\overline{y}}_{1} is zero of (7) of multiplicity three. In view of Lemmas 6 and 7, from [27] we have that {\tilde{x}}^{\prime}({\overline{y}}_{1})=0. The rest of the proof is similar to that in Theorem 12 and we skip it. □
4.1 Periodtwo solution
Let
{T}^{2}(x,y)=T(T(x,y))=(F(x,y),G(x,y)),
where
F(x,y)=\frac{{b}_{1}^{3}{x}^{8}}{{({A}_{1}+{y}^{2})}^{2}(2{a}_{2}{c}_{2}{y}^{2}+{a}_{2}^{2}+{A}_{1}{x}^{4}+{c}_{2}^{2}{y}^{4})}
and
G(x,y)=\frac{{({A}_{1}+{y}^{2})}^{2}(2{a}_{2}{c}_{2}^{2}{y}^{2}+{a}_{2}^{2}{c}_{2}+{a}_{2}{x}^{4}+{c}_{2}^{3}{y}^{4})}{{b}_{1}^{2}{x}^{8}}.
Periodtwo solution \{(\mathrm{\Phi},\mathrm{\Psi}),T(\mathrm{\Phi},\mathrm{\Psi})\} satisfies the system
F(\mathrm{\Phi},\mathrm{\Psi})=\mathrm{\Phi},\phantom{\rule{2em}{0ex}}G(\mathrm{\Phi},\mathrm{\Psi})=\mathrm{\Psi},
which is equivalent to
\begin{array}{r}{b}_{1}^{3}{\mathrm{\Phi}}^{7}{A}_{1}{\mathrm{\Phi}}^{4}{({A}_{1}+{\mathrm{\Psi}}^{2})}^{2}{({A}_{1}+{\mathrm{\Psi}}^{2})}^{2}{({a}_{2}+{c}_{2}{\mathrm{\Psi}}^{2})}^{2}=0,\\ {b}_{1}^{2}\mathrm{\Psi}{\mathrm{\Phi}}^{8}{a}_{2}{\mathrm{\Phi}}^{4}{({A}_{1}+{\mathrm{\Psi}}^{2})}^{2}{c}_{2}{({A}_{1}+{\mathrm{\Psi}}^{2})}^{2}{({a}_{2}+{c}_{2}{\mathrm{\Psi}}^{2})}^{2}=0.\end{array}
(22)
The Jacobian matrix of the map {T}^{2} at (x,y) has the form
{J}_{{T}^{2}}(x,y)=\left(\begin{array}{cc}\frac{4{x}^{7}{b}_{1}^{3}({A}_{1}{x}^{4}+2{({c}_{2}{y}^{2}+{a}_{2})}^{2})}{{({y}^{2}+{A}_{1})}^{2}{({A}_{1}{x}^{4}+{({c}_{2}{y}^{2}+{a}_{2})}^{2})}^{2}}& \frac{4{x}^{8}y{b}_{1}^{3}({A}_{1}{x}^{4}+({c}_{2}{y}^{2}+{a}_{2})({a}_{2}+(2{y}^{2}+{A}_{1}){c}_{2}))}{{({y}^{2}+{A}_{1})}^{3}{({A}_{1}{x}^{4}+{({c}_{2}{y}^{2}+{a}_{2})}^{2})}^{2}}\\ \frac{4{({y}^{2}+{A}_{1})}^{2}(2{c}_{2}^{3}{y}^{4}+2{a}_{2}^{2}{c}_{2}+{a}_{2}({x}^{4}+4{y}^{2}{c}_{2}^{2}))}{{x}^{9}{b}_{1}^{2}}& \frac{4y({y}^{2}+{A}_{1})({y}^{2}(2{y}^{2}+{A}_{1}){c}_{2}^{3}+{a}_{2}^{2}{c}_{2}+{a}_{2}({x}^{4}+(3{y}^{2}+{A}_{1}){c}_{2}^{2}))}{{x}^{8}{b}_{1}^{2}}\end{array}\right).
(23)
The determinant of (23) at (x,y) is given by
det{J}_{{T}^{2}}(x,y)=\frac{16{b}_{1}{x}^{3}y{({a}_{2}{A}_{1}{c}_{2})}^{2}({a}_{2}+{c}_{2}{y}^{2})}{({A}_{1}+{y}^{2}){({({a}_{2}+{c}_{2}{y}^{2})}^{2}+{A}_{1}{x}^{4})}^{2}}.
The trace of (23) at (x,y) is given by
\begin{array}{rl}tr{J}_{{T}^{2}}(x,y)=& \frac{4(\frac{{b}_{1}^{5}{x}^{15}(2{({a}_{2}+{c}_{2}{y}^{2})}^{2}+{A}_{1}{x}^{4})}{{({({a}_{2}+{c}_{2}{y}^{2})}^{2}+{A}_{1}{x}^{4})}^{2}}+{a}_{2}y{({A}_{1}+{y}^{2})}^{3}({c}_{2}^{2}({A}_{1}+3{y}^{2})+{x}^{4}))}{{b}_{1}^{2}{x}^{8}{({A}_{1}+{y}^{2})}^{2}}\\ +\frac{4({a}_{2}^{2}{c}_{2}y{({A}_{1}+{y}^{2})}^{3}+{c}_{2}^{3}{y}^{3}({A}_{1}+2{y}^{2}){({A}_{1}+{y}^{2})}^{3})}{{b}_{1}^{2}{x}^{8}{({A}_{1}+{y}^{2})}^{2}}.\end{array}
(24)
Lemma 6 Let {\mathcal{C}}_{F}:=\{(x,y):F(x,y)=x\} and {\mathcal{C}}_{G}:=\{(x,y):G(x,y)=y\} be the periodtwo curves, that is, the curves the intersection of which is a periodtwo solution. Then, for all y>0, there exist exactly one {x}_{F}(y)>0 and exactly one {x}_{G}(y)>0 such that F({x}_{F}(y),y)=x and G({x}_{G}(y),y)=y. Furthermore, {x}_{F}(y) and {x}_{G}(y) are continuous functions and {x}_{F}^{\prime}(y)>0.
Proof Since F(x,y)=x and G(x,y)=y if and only if
\begin{array}{c}{b}_{1}^{3}{x}^{7}{A}_{1}{x}^{4}{({A}_{1}+{y}^{2})}^{2}{({A}_{1}+{y}^{2})}^{2}{({a}_{2}+{c}_{2}{y}^{2})}^{2}=0,\hfill \\ {b}_{1}^{2}y{x}^{8}{a}_{2}{x}^{4}{({A}_{1}+{y}^{2})}^{2}{c}_{2}{({A}_{1}+{y}^{2})}^{2}{({a}_{2}+{c}_{2}{y}^{2})}^{2}=0,\hfill \end{array}
respectively, in view of Descartes’ rule of signs, we have that for all y>0 there exist exactly one {x}_{F}(y)>0 and exactly one {x}_{G}(y)>0 such that F({x}_{F}(y),y)=x and G({x}_{G}(y),y)=y. Taking derivatives of F(x,y)=x with respect to y, we get
{x}_{F}^{\prime}(y)=\frac{{F}_{y}^{\prime}(x,y)}{1{F}_{x}^{\prime}(x,y)}.
From F(x,y)=x we have that {({A}_{1}+{y}^{2})}^{2}=\frac{{b}_{1}^{3}{x}^{7}}{{({a}_{2}+{c}_{2}{y}^{2})}^{2}+{A}_{1}{x}^{4}}, which implies
{F}_{x}^{\prime}(x,y)=\frac{4{x}^{7}{b}_{1}^{3}({A}_{1}{x}^{4}+2{({c}_{2}{y}^{2}+{a}_{2})}^{2})}{{({y}^{2}+{A}_{1})}^{2}{({A}_{1}{x}^{4}+{({c}_{2}{y}^{2}+{a}_{2})}^{2})}^{2}}=8\frac{4{A}_{1}{x}^{4}}{{({a}_{2}+{c}_{2}{y}^{2})}^{2}+{A}_{1}{x}^{4}}>4.
Since {F}_{y}^{\prime}(x,y)<0, we get {x}_{F}^{\prime}(y)>0. □
Theorem 14 If {a}_{2}\le {A}_{1}{c}_{2}, then T has no minimal periodtwo solution. If {a}_{2}>{A}_{1}{c}_{2} and T has a minimal periodtwo solution \{(\mathrm{\Phi},\mathrm{\Psi}),T(\mathrm{\Phi},\mathrm{\Psi})\}, then \{(\mathrm{\Phi},\mathrm{\Psi}),T(\mathrm{\Phi},\mathrm{\Psi})\} is unstable. If {\mu}_{1} and {\mu}_{2} ({\mu}_{1}<{\mu}_{2}) are the eigenvalues of {J}_{{T}^{2}}(\mathrm{\Phi},\mathrm{\Psi}), then {\mu}_{1}>0 and {\mu}_{2}>1. All periodtwo solutions are ordered with respect to the northeast ordering.
Proof If {a}_{2}\le {A}_{1}{c}_{2}, the statement follows from Lemma 2. If {a}_{2}>{A}_{1}{c}_{2}, then from the first equation of (22) we have that
{({A}_{1}+{\mathrm{\Psi}}^{2})}^{2}=\frac{{b}_{1}^{3}{\mathrm{\Phi}}^{7}}{{({a}_{2}+{c}_{2}{\mathrm{\Psi}}^{2})}^{2}+{A}_{1}{\mathrm{\Phi}}^{4}}.
By substituting this into (24) we obtain
\begin{array}{rl}tr{J}_{{T}^{2}}(\mathrm{\Phi},\mathrm{\Psi})=& \frac{4\mathrm{\Psi}({A}_{1}+{\mathrm{\Psi}}^{2})({a}_{2}({c}_{2}^{2}({A}_{1}+3{\mathrm{\Psi}}^{2})+{\mathrm{\Phi}}^{4})+{a}_{2}^{2}{c}_{2}+{c}_{2}^{3}{\mathrm{\Psi}}^{2}({A}_{1}+2{\mathrm{\Psi}}^{2}))}{{b}_{1}^{2}{\mathrm{\Phi}}^{8}}\\ +8\frac{4{A}_{1}{\mathrm{\Phi}}^{4}}{{({a}_{2}+{c}_{2}{\mathrm{\Psi}}^{2})}^{2}+{A}_{1}{\mathrm{\Phi}}^{4}}>4.\end{array}
The rest of the proof follows from the fact that tr{J}_{{T}^{2}}(\mathrm{\Phi},\mathrm{\Psi})={\mu}_{1}+{\mu}_{2}>4, det{J}_{{T}^{2}}(\mathrm{\Phi},\mathrm{\Psi})={\mu}_{1}{\mu}_{2}>0 and Lemma 6. □
Theorem 15 If the map T has a minimal periodtwo solution \{({\mathrm{\Phi}}_{1},{\mathrm{\Psi}}_{1}),T({\mathrm{\Phi}}_{1},{\mathrm{\Psi}}_{1})\}, which is nonhyperbolic, then D(p)=0, where D(p) is the discriminant of the polynomial
p(y):={p}_{22}{y}^{22}+{p}_{21}{y}^{21}+\cdots +{p}_{1}y+{p}_{0},
where the coefficients {p}_{i}, i=0,\dots ,22, are in the Appendix. If \{({\mathrm{\Phi}}_{1},{\mathrm{\Psi}}_{1}),T({\mathrm{\Phi}}_{1},{\mathrm{\Psi}}_{1})\} and \{({\mathrm{\Phi}}_{2},{\mathrm{\Psi}}_{2}),T({\mathrm{\Phi}}_{2},{\mathrm{\Psi}}_{2})\} are two minimal periodtwo solutions such that T has no other minimal periodtwo solutions in \u301a({\mathrm{\Phi}}_{1},{\mathrm{\Psi}}_{1}),({\mathrm{\Phi}}_{2},{\mathrm{\Psi}}_{2})\u301b=\{(x,y):({\mathrm{\Phi}}_{1},{\mathrm{\Psi}}_{1}){\u2aaf}_{ne}(x,y){\u2aaf}_{ne}({\mathrm{\Phi}}_{2},{\mathrm{\Psi}}_{2})\} and D(p)\ne 0, then one of them is a saddle point and the other is a repeller.
Proof Periodtwo solution curves {\mathcal{C}}_{F}=\{(x,y)\in \mathcal{R}:\tilde{F}(x,y)=0\} and {\mathcal{C}}_{G}=\{(x,y)\in \mathcal{R}:\tilde{G}(x,y)=0\}, where
\begin{array}{c}\tilde{F}(x,y)={b}_{1}^{3}{x}^{7}{A}_{1}{x}^{4}{({A}_{1}+{y}^{2})}^{2}{({A}_{1}+{y}^{2})}^{2}{({a}_{2}+{c}_{2}{y}^{2})}^{2}\phantom{\rule{1em}{0ex}}\text{and}\hfill \\ \tilde{G}(x,y)={b}_{1}^{2}y{x}^{8}{a}_{2}{x}^{4}{({A}_{1}+{y}^{2})}^{2}{c}_{2}{({A}_{1}+{y}^{2})}^{2}{({a}_{2}+{c}_{2}{y}^{2})}^{2},\hfill \end{array}
are algebraic curves. By using software Mathematica, one can see that the resultant of the polynomials \tilde{F}(x,y) and \tilde{G}(x,y) in variable x is given by
\begin{array}{rl}R(\tilde{F},\tilde{G})& ={b}_{1}^{6}{({A}_{1}+{y}^{2})}^{14}{({a}_{2}+{c}_{2}{y}^{2})}^{8}({a}_{2}{b}_{1}^{2}+2{A}_{1}{y}^{3}+{A}_{1}^{2}y{b}_{1}^{2}{c}_{2}{y}^{2}+{y}^{5})p(y)\\ ={b}_{1}^{6}{({A}_{1}+{y}^{2})}^{14}{({a}_{2}+{c}_{2}{y}^{2})}^{8}\tilde{f}(y)p(y).\end{array}
Suppose that \{({\mathrm{\Phi}}_{1},{\mathrm{\Psi}}_{1}),T({\mathrm{\Phi}}_{1},{\mathrm{\Psi}}_{1})\} is a nonhyperbolic minimal periodtwo solution. This implies that \tilde{F}({\mathrm{\Phi}}_{1},{\mathrm{\Psi}}_{1})=0, \tilde{G}({\mathrm{\Phi}}_{1},{\mathrm{\Psi}}_{1})=0. By Theorem 9.3 [28], \tilde{F} and \tilde{G} have a common nonconstant factor if and only R(\tilde{F},\tilde{G})=0, which implies that system \tilde{F}(x,y)=0, \tilde{G}(x,y)=0 has a solution if and only if R(\tilde{F},\tilde{G})=0. Since \tilde{f}({\mathrm{\Psi}}_{1})\ne 0, it must be p({\mathrm{\Psi}}_{1})=0. Similarly as in Lemma 4, one can see that
\begin{array}{rl}{x}_{F}^{\prime}({\mathrm{\Psi}}_{1}){x}_{G}^{\prime}({\mathrm{\Psi}}_{1})& =\frac{{F}_{y}^{\prime}({\mathrm{\Phi}}_{1},{\mathrm{\Psi}}_{1})}{1{F}_{x}^{\prime}({\mathrm{\Phi}}_{1},{\mathrm{\Psi}}_{1})}\frac{1{G}_{y}^{\prime}({\mathrm{\Phi}}_{1},{\mathrm{\Psi}}_{1})}{{G}_{x}^{\prime}({\mathrm{\Phi}}_{1},{\mathrm{\Psi}}_{1})}\\ =\frac{{f}_{1}}{1{e}_{1}}\frac{1{h}_{1}}{{g}_{1}}=\frac{1+({e}_{1}+{g}_{1})({e}_{1}{h}_{1}{f}_{1}{g}_{1})}{{g}_{1}(1{e}_{1})}\\ =\frac{{p}_{1}(1)}{{g}_{1}(1{e}_{1})}=\frac{(1{\mu}_{1})(1{\mu}_{2})}{{g}_{1}({e}_{1}1)},\end{array}
where {p}_{1}(\mu ) is the characteristic equation of the matrix
{J}_{{T}^{2}}({\mathrm{\Phi}}_{1},{\mathrm{\Psi}}_{1})=\left(\begin{array}{cc}{e}_{1}& {f}_{1}\\ {g}_{1}& {h}_{1}\end{array}\right).
From Theorem 14 we have that 0<{\mu}_{1}<{\mu}_{2} and {\mu}_{2}>1. Since \{({\mathrm{\Phi}}_{1},{\mathrm{\Psi}}_{1}),T({\mathrm{\Phi}}_{1},{\mathrm{\Psi}}_{1})\} is nonhyperbolic, we obtain that {\mu}_{1}=1, from which it follows that {x}_{F}^{\prime}({\mathrm{\Psi}}_{1}){x}_{G}^{\prime}({\mathrm{\Psi}}_{1})=0. Since R(\tilde{F},\tilde{G})\not\equiv 0, we have that {\mathcal{C}}_{F} and {\mathcal{C}}_{G} have no common component. By Lemmas 6 and 7, from [27], the curves {\mathcal{C}}_{F} and {\mathcal{C}}_{G} intersect transversally at ({\mathrm{\Phi}}_{1},{\mathrm{\Psi}}_{1}) (i.e., {y}_{\tilde{F}}^{\prime}({\mathrm{\Psi}}_{1}){y}_{\tilde{G}}^{\prime}({\mathrm{\Psi}}_{1})\ne 0) if and only if {\mathrm{\Psi}}_{1} is zero of p(y) of multiplicity one. By Theorem 9.4 [28], p(y) has zeros of multiplicity greater than one if and only if the discriminant D(p) of the polynomial p(y) is equal to zero, which proves the first statement of the lemma.
Assume that \{({\mathrm{\Phi}}_{1},{\mathrm{\Psi}}_{1}),T({\mathrm{\Phi}}_{1},{\mathrm{\Psi}}_{1})\} and \{({\mathrm{\Phi}}_{2},{\mathrm{\Psi}}_{2}),T({\mathrm{\Phi}}_{2},{\mathrm{\Psi}}_{2})\} are two minimal periodtwo solutions such that T has no other minimal periodtwo solutions in \u301a({\mathrm{\Phi}}_{1},{\mathrm{\Psi}}_{1}),({\mathrm{\Phi}}_{2},{\mathrm{\Psi}}_{2})\u301b=\{(x,y):({\mathrm{\Phi}}_{1},{\mathrm{\Psi}}_{1}){\u2aaf}_{ne}(x,y){\u2aaf}_{ne}({\mathrm{\Phi}}_{2},{\mathrm{\Psi}}_{2})\} and D(p)\ne 0. From the previous discussion we have {x}_{F}^{\prime}({\mathrm{\Psi}}_{i}){x}_{G}^{\prime}({\mathrm{\Psi}}_{i})\ne 0, i=1,2. Since {x}_{F}({\mathrm{\Psi}}_{i}){x}_{G}({\mathrm{\Psi}}_{i})=0, i=1,2, it follows that ({x}_{F}^{\prime}({\mathrm{\Psi}}_{1}){x}_{G}^{\prime}({\mathrm{\Psi}}_{1}))({x}_{F}^{\prime}({\mathrm{\Psi}}_{2}){x}_{G}^{\prime}({\mathrm{\Psi}}_{2}))<0. Indeed assume, for example, that {x}_{F}^{\prime}({\mathrm{\Psi}}_{1}){x}_{G}^{\prime}({\mathrm{\Psi}}_{1})<0 and {x}_{F}^{\prime}({\mathrm{\Psi}}_{2}){x}_{G}^{\prime}({\mathrm{\Psi}}_{2})<0. Then there exists \u03f5>0 such that {x}_{F}(y){x}_{G}(y)<0 for y\in ({\mathrm{\Psi}}_{1},{\mathrm{\Psi}}_{1}+\u03f5) and {x}_{F}(y){x}_{G}(y)>0 for y\in ({\mathrm{\Psi}}_{2}\u03f5,{\mathrm{\Psi}}_{2}). Since {x}_{F}(y){x}_{G}(y) is a continuous function, this implies that there exists \mathrm{\Psi}\in ({\mathrm{\Psi}}_{1},{\mathrm{\Psi}}_{2}) such that {x}_{F}(\mathrm{\Psi}){x}_{G}(\mathrm{\Psi})=0, which is a contradiction. The rest of the proof follows from the fact that {e}_{i}>4 and {g}_{i}<0, i=1,2. □
Notice that
D(p)=\frac{1}{{p}_{22}}R(p,{p}^{\prime}),
where R(p,{p}^{\prime})=detSyl(p,{p}^{\prime}), the determinant of the Sylvester matrix Syl(p,{p}^{\prime}), see [28, 29], and
Syl(p,{p}^{\prime})=\left(\begin{array}{cccccccc}{p}_{22}& {p}_{21}& \cdots & {p}_{1}& {p}_{0}& 0& \cdots & 0\\ 0& {p}_{22}& {p}_{21}& \cdots & {p}_{1}& {p}_{0}& \cdots & 0\\ \vdots \\ 0& \cdots & {p}_{22}& {p}_{21}& {p}_{20}& \cdots & {p}_{1}& {p}_{0}\\ 22{p}_{21}& 21{p}_{20}& \cdots & {p}_{1}& 0& 0& \cdots & 0\\ 0& 22{p}_{21}& 21{p}_{20}& \cdots & {p}_{1}& 0& \cdots & 0\\ \vdots \\ 0& 0& \cdots & 22{p}_{21}& 21{p}_{20}& 20{p}_{19}& \cdots & {p}_{1}\end{array}\right).
Theorem 16 If {a}_{2}>{A}_{1}{c}_{2} and \mathrm{\Gamma}(\overline{y})<0, then T has one equilibrium point E(\overline{x},\overline{y}), which is a repeller, and there exists at least one minimal periodtwo solution \{(\psi ,\varphi ),T(\psi ,\varphi )\} which is nonhyperbolic or a saddle point. If T has no minimal periodtwo solutions which are nonhyperbolic, then (\psi ,\varphi ){\ll}_{ne}E{\ll}_{ne}T(\psi ,\varphi ).
Proof By Theorem 10 we have that T has one equilibrium point E(\overline{x},\overline{y}), which is a repeller. This and Lemma 2 imply that {T}^{n}({x}_{0},{y}_{0}) is asymptotic to either (0,\mathrm{\infty}) or (\mathrm{\infty},0), or a minimal periodtwo solution, for all ({x}_{0},{y}_{0})\in \mathcal{R}. Let \mathcal{B}(0,\mathrm{\infty}) be the basin of attraction of (0,\mathrm{\infty}), and let \mathcal{B}(\mathrm{\infty},0) be the basin of attraction of (\mathrm{\infty},0). By using Theorem 9 one can prove that int({Q}_{2}(E))\subset \mathcal{B}(0,\mathrm{\infty}) and int({Q}_{4}(E))\subset \mathcal{B}(\mathrm{\infty},0). Let {\mathcal{S}}_{1} denote the boundary of \mathcal{B}(\mathrm{\infty},0) considered as a subset of {Q}_{1}(E), and let {\mathcal{S}}_{2} denote the boundary of \mathcal{B}(\mathrm{\infty},0) considered as a subset of {Q}_{3}(E). It is easy to see that E\in {\mathcal{S}}_{1}, E\in {\mathcal{S}}_{2} and T(\mathcal{R})\subset int(\mathcal{R}). Now we prove the following claim.
Claim 1 Let {\mathcal{S}}_{1} and {\mathcal{S}}_{2} be the sets defined as above. Then

(a)
If ({x}_{0},{y}_{0})\in \mathcal{B}(\mathrm{\infty},0), then ({x}_{1},{y}_{1})\in \mathcal{B}(\mathrm{\infty},0) for all ({x}_{0},{y}_{0}){\u2aaf}_{se}({x}_{1},{y}_{1}).

(b)
If ({x}_{0},{y}_{0})\in {\mathcal{S}}_{1}\cup {\mathcal{S}}_{2}, then ({x}_{1},{y}_{1})\in int(\mathcal{B})(\mathrm{\infty},0) for all ({x}_{0},{y}_{0}){\ll}_{se}({x}_{1},{y}_{1}).

(c)
{\mathcal{S}}_{1}\cap int({Q}_{1}(E))\ne \mathrm{\varnothing} and {\mathcal{S}}_{2}\cap int({Q}_{3}(E))\ne \mathrm{\varnothing}.

(d)
T({\mathcal{S}}_{1}\cup {\mathcal{S}}_{2})\subseteq {\mathcal{S}}_{1}\cup {\mathcal{S}}_{2}.

(e)
({x}_{0},{y}_{0}),({x}_{1},{y}_{1})\in {\mathcal{S}}_{1}\cup {\mathcal{S}}_{2}\Rightarrow ({x}_{0},{y}_{0}){\ll}_{ne}({x}_{1},{y}_{1}) or ({x}_{1},{y}_{1}){\ll}_{ne}({x}_{0},{y}_{0}).
Proof

(a)
The statement follows from {T}^{n}({x}_{0},{y}_{0}){\u2aaf}_{se}{T}^{n}({x}_{1},{y}_{1}){\u2aaf}_{se}(\mathrm{\infty},0) and {T}^{n}({x}_{0},{y}_{0})\to (\mathrm{\infty},0) as n\to \mathrm{\infty}.

(b)
The claim (b) follows from the observation that there exists a ball centered at ({x}_{0},{y}_{0}) with the property that all its points (x,y) satisfy (x,y){\ll}_{se}({x}_{1},{y}_{1}). But one of these points necessarily lies in \mathcal{B}(\mathrm{\infty},0), so by (a) there exists ({x}_{1},{y}_{1})\in \mathcal{B}(\mathrm{\infty},0). Furthermore, there exists a ball centered at ({x}_{1},{y}_{1}) with the property that all its points (x,y) satisfy (x,y)\in \mathcal{B}(\mathrm{\infty},0), which implies ({x}_{1},{y}_{1})\in int(\mathcal{B})(\mathrm{\infty},0).

(c)
Take {y}^{\prime}>\overline{y} arbitrary (but fixed). Since T is strongly competitive, we have T(\overline{x},{y}^{\prime}){\ll}_{se}T(\overline{x},\overline{y}), which implies T(\overline{x},{y}^{\prime})\in int({Q}_{2}(E)). This implies that there exists a ball {B}_{\epsilon}(T(\overline{x},{y}^{\prime})) with the property {B}_{\epsilon}(T(\overline{x},{y}^{\prime}))\subset int({Q}_{2}(E)). Since T is a continuous map on a set {\mathbb{R}}_{+}^{2}\setminus \{(0,y):y\ge 0\}, then there exists a ball {B}_{{\delta}_{1}}(\overline{x},{y}^{\prime}) such that T({B}_{{\delta}_{1}}(\overline{x},{y}^{\prime}))\subset {B}_{\epsilon}(T(\overline{x},{y}^{\prime}))\subset int({Q}_{2}(E)), which implies {T}^{n}(x,y)\to (0,\mathrm{\infty}) as n\to \mathrm{\infty} for all (x,y)\in {B}_{{\delta}_{1}}(\overline{x},{y}^{\prime}). Similarly, one can prove that then there exists a ball {B}_{{\delta}_{2}}(\overline{x}+{\delta}_{1}/2,\overline{y}) such that {T}^{n}(x,y)\to (\mathrm{\infty},0) as n\to \mathrm{\infty} for all (x,y)\in {B}_{{\delta}_{2}}(\overline{x}+{\delta}_{1}/2,\overline{y}). Let {y}^{\u2033}=sup\{y:{lim}_{n\to \mathrm{\infty}}{T}^{n}(\overline{x}+{\delta}_{1}/2,y)=(\mathrm{\infty},0)\}. It is easy to see that (\overline{x}+{\delta}_{1}/2,{y}^{\u2033})\in {\mathcal{S}}_{1}\cap int({Q}_{1}(E)). The assertion concerning {\mathcal{S}}_{2} is proved in a similar fashion.

(d)
Take (x,y)\in {\mathcal{S}}_{1}\cup {\mathcal{S}}_{2}. Assume that T(x,y)\notin {\mathcal{S}}_{1}\cup {\mathcal{S}}_{2}. Since {\mathcal{S}}_{1}\cup {\mathcal{S}}_{2}=\partial \mathcal{B}(\mathrm{\infty},0)=\overline{\mathcal{B}(\mathrm{\infty},0)}\setminus int(\mathcal{B}(\mathrm{\infty},0)), then either T(x,y)\in int(\mathcal{B}(\mathrm{\infty},0)) or T(x,y)\notin \overline{\mathcal{B}(\mathrm{\infty},0)}. Assume that T(x,y)\in int(\mathcal{B}(\mathrm{\infty},0)). This implies that there exists a ball {B}_{\epsilon}(T(x,y)) with the property {B}_{\epsilon}(T(x,y))\subset int(\mathcal{B}(\mathrm{\infty},0)). Since T is a continuous map on the set {\mathbb{R}}_{+}^{2}\setminus \{(0,y):y\ge 0\}, then there exists a ball {B}_{\delta}(x,y), \delta >0 such that T({B}_{\delta}(x,y))\subset {B}_{\epsilon}(T(x,y)), which implies {B}_{\delta}(x,y)\subset \mathcal{B}(\mathrm{\infty},0). This is in contradiction with (x,y)\in {\mathcal{S}}_{1}\cup {\mathcal{S}}_{2}=\partial \mathcal{B}(\mathrm{\infty},0). Hence T(x,y)\in {\mathcal{S}}_{1}\cup {\mathcal{S}}_{2} in this case. Similarly, one can prove that T(x,y)\in {\mathcal{S}}_{1}\cup {\mathcal{S}}_{2} if T(x,y)\notin \overline{\mathcal{B}(\mathrm{\infty},0)}. This implies that T({\mathcal{S}}_{1}\cup {\mathcal{S}}_{2})\subseteq ({\mathcal{S}}_{1}\cup {\mathcal{S}}_{2}).

(e)
Assume that ({x}_{0},{y}_{0}),({x}_{1},{y}_{1})\in {\mathcal{S}}_{1}\cup {\mathcal{S}}_{2}\Rightarrow ({x}_{0},{y}_{0}){\u2aaf}_{se}({x}_{1},{y}_{1}) and ({x}_{0},{y}_{0})\ne ({x}_{1},{y}_{1}). Since T is strongly competitive, we get T({x}_{0},{y}_{0}){\ll}_{ne}T({x}_{1},{y}_{1}). This contradicts (e) and (b), which completes the proof. □
In view of Claim 1, we have that ({\mathcal{S}}_{1}\cup {\mathcal{S}}_{2},{\ll}_{ne}) is a totally ordered set which is invariant under T. If ({x}_{0},{y}_{0})\in {\mathcal{S}}_{1}\cup {\mathcal{S}}_{2}, then \{{T}^{(2n)}({x}_{0},{y}_{0})\} is eventually componentwise monotone. Then there exists a minimal periodtwo solution \{(\mathrm{\Phi},\mathrm{\Psi}),T(\mathrm{\Phi},\mathrm{\Psi})\}\in {\mathcal{S}}_{1}\cup {\mathcal{S}}_{2}\subset {Q}_{1}(E)\cup {Q}_{3}(E) such that {T}^{(2n)}({x}_{0},{y}_{0})\to (\mathrm{\Phi},\mathrm{\Psi}) as n\to \mathrm{\infty}. By Theorem 14, \{(\mathrm{\Phi},\mathrm{\Psi}),T(\mathrm{\Phi},\mathrm{\Psi})\} is a nonhyperbolic or a saddle point. Assume that T has no minimal periodtwo solutions which are nonhyperbolic points and, for example, that (\mathrm{\Phi},\mathrm{\Psi}),T(\mathrm{\Phi},\mathrm{\Psi})\in {\mathcal{S}}_{1} such that E{\ll}_{ne}(\mathrm{\Phi},\mathrm{\Psi}){\ll}_{ne}T(\mathrm{\Phi},\mathrm{\Psi}). Since \{(\mathrm{\Phi},\mathrm{\Psi}),T(\mathrm{\Phi},\mathrm{\Psi})\} is a saddle point, in view of Theorems 6, 7 and 8, we have that the global stable manifolds {\mathcal{W}}^{s}(\{(\mathrm{\Phi},\mathrm{\Psi}),T(\mathrm{\Phi},\mathrm{\Psi})\}) are the union of two curves {\mathcal{W}}^{s}(\mathrm{\Phi},\mathrm{\Psi}) and {\mathcal{W}}^{s}(T(\mathrm{\Phi},\mathrm{\Psi})) whose endpoints are repeller points such that T({\mathcal{W}}^{s}(\mathrm{\Phi},\mathrm{\Psi}))={\mathcal{W}}^{s}(T(\mathrm{\Phi},\mathrm{\Psi})) and E{\ll}_{ne}\overline{{\mathcal{W}}^{s}(\mathrm{\Phi},\mathrm{\Psi})}{\ll}_{ne}\overline{{\mathcal{W}}^{s}(T(\mathrm{\Phi},\mathrm{\Psi}))}. If {P}_{1} and {P}_{2} ({P}_{1}{\u2aaf}_{ne}{P}_{2}) are endpoints of {\mathcal{W}}^{s}(\mathrm{\Phi},\mathrm{\Psi}), then T({P}_{1}) and T({P}_{2}) are endpoints of T({\mathcal{W}}^{s}(\mathrm{\Phi},\mathrm{\Psi})) and either T({P}_{1}){\u2aaf}_{ne}T({P}_{2}) or T({P}_{2}){\u2aaf}_{ne}T({P}_{1}). Assume, for example, that {P}_{1}{\ll}_{ne}{\mathcal{W}}^{s}(\mathrm{\Phi},\mathrm{\Psi}){\ll}_{ne}{P}_{2}{\ll}_{ne}T({P}_{2}){\ll}_{ne}{\mathcal{W}}^{s}(T(\mathrm{\Phi},\mathrm{\Psi})){\ll}_{ne}T({P}_{1}). By Theorem 15 between two repellers {P}_{2} and T({P}_{2}), there exists a saddle point {S}_{1} where its stable manifold is the union of two invariant curves {\mathcal{W}}^{s}({S}_{1}) and {\mathcal{W}}^{s}(T({S}_{1})) whose endpoints are repellers such that {P}_{2}{\ll}_{ne}\overline{{\mathcal{W}}^{s}({S}_{1})}{\ll}_{ne}T({P}_{2}). Continuing in this way, we obtain that T has infinitely many minimal periodtwo solutions \{{P}_{i},T({P}_{i})\}, which is in contradiction with the fact that T has at most eleven minimal periodtwo solutions. Hence (\mathrm{\Phi},\mathrm{\Psi}){\ll}_{ne}E{\ll}_{ne}T(\mathrm{\Phi},\mathrm{\Psi}). □
Corollary 2 Assume that {a}_{2}>{A}_{1}{c}_{2}, \mathrm{\Gamma}(\overline{y})<0 and D(p)\ne 0. Then there exists one equilibrium point E which is a repeller. Further, the set int({Q}_{1}(E))\cup int({Q}_{3}(E)) contains an odd number of minimal periodtwo solutions \{({\mathrm{\Phi}}_{i},{\mathrm{\Psi}}_{i}),({\tilde{\mathrm{\Phi}}}_{i},{\tilde{\mathrm{\Psi}}}_{i})\}, i=1,\dots ,2n+1, such that ({\mathrm{\Phi}}_{i+1},{\mathrm{\Psi}}_{i+1}){\ll}_{ne}({\mathrm{\Phi}}_{i},{\mathrm{\Psi}}_{i}){\ll}_{ne}E and E{\ll}_{ne}({\tilde{\mathrm{\Phi}}}_{i},{\tilde{\mathrm{\Psi}}}_{i}){\ll}_{ne}({\tilde{\mathrm{\Phi}}}_{i+1},{\tilde{\mathrm{\Psi}}}_{i+1}), where ({\tilde{\mathrm{\Phi}}}_{i},{\tilde{\mathrm{\Psi}}}_{i})=T({\mathrm{\Phi}}_{i},{\mathrm{\Psi}}_{i}). Furthermore, odd indexed periodtwo points are saddles and even indexed periodtwo points are repellers.
Proof By Theorem 10 we have that T has one equilibrium point E(\overline{x},\overline{y}), which is a repeller. By Theorem 15 all minimal periodtwo solutions are hyperbolic and the number of minimal periodtwo solutions is finite. In view of Claim 1, T has at least one minimal periodtwo solution which is a saddle point. Let \{{P}_{1},T({P}_{1})\} be a minimal periodtwo solution which is a saddle point such that {P}_{1}{\ll}_{ne}E{\ll}_{ne}T({P}_{1}) and T has no minimal periodtwo solutions in \u301aE,T({P}_{1})\u301b and \u301a{P}_{1},E\u301b. Such a minimal periodtwo solution exists in view of Theorem 15. The map {T}^{2} satisfies all conditions of Theorems 6, 7 and 8, which yields the existence of the global stable manifolds {\mathcal{W}}^{s}(\{{P}_{1},{\tilde{P}}_{1}\}) which are the union of two curves {\mathcal{W}}^{s}({P}_{1}) and {\mathcal{W}}^{s}({\tilde{P}}_{1}) that have a common endpoint E. If T has minimal periodtwo solutions in int({Q}_{1}(T({P}_{1})))\cup int({Q}_{3}({P}_{1})), let \{{P}_{2},{\tilde{P}}_{2}\} ({P}_{2}{\ll}_{ne}{\tilde{P}}_{2}) denote minimal periodtwo solutions such that T has no other minimal periodtwo solutions in \u301aT({P}_{1}),T({P}_{2})\u301b and \u301a{P}_{2},{P}_{1}\u301b. Then {\mathcal{W}}^{s}({P}_{1}) has the second endpoint at {P}_{2} and {\mathcal{W}}^{s}(T({P}_{1})) has the second endpoint at T({P}_{2}) and {P}_{2}{\ll}_{ne}{P}_{1}{\ll}_{ne}E{\ll}_{ne}T({P}_{1}){\ll}_{ne}T({P}_{2}). Furthermore, a minimal periodtwo solution \{{P}_{2},T({P}_{2})\} is a repeller. Similarly as in Theorem 16, one can prove that int({Q}_{1}(T({P}_{2})))\cup int({Q}_{3}({P}_{2})) contains at least one minimal periodtwo solution which is a saddle point. Since the number of minimal periodtwo solutions is finite, continuing in this way, we will end with a minimal periodtwo solution which is a saddle point, from which the proof follows. □
Corollary 3 If {a}_{2}>{A}_{1}{c}_{2} and \mathrm{\Gamma}(\overline{y})>0, then there exists one equilibrium point E which is a saddle point. If D(p)\ne 0, then int({Q}_{1}(E))\cup int({Q}_{3}(E)) contains an even number of minimal periodtwo solutions \{({\mathrm{\Phi}}_{i},{\mathrm{\Psi}}_{i}),({\tilde{\mathrm{\Phi}}}_{i},{\tilde{\mathrm{\Psi}}}_{i})\}, i=1,\dots ,2n, such that ({\mathrm{\Phi}}_{i+1},{\mathrm{\Psi}}_{i+1}){\ll}_{ne}({\mathrm{\Phi}}_{i},{\mathrm{\Psi}}_{i}){\ll}_{ne}E and E{\ll}_{ne}({\tilde{\mathrm{\Phi}}}_{i},{\tilde{\mathrm{\Psi}}}_{i}){\ll}_{ne}({\tilde{\mathrm{\Phi}}}_{i+1},{\tilde{\mathrm{\Psi}}}_{i+1}) and ({\tilde{\mathrm{\Phi}}}_{i},{\tilde{\mathrm{\Psi}}}_{i})=T({\mathrm{\Phi}}_{i},{\mathrm{\Psi}}_{i}). Furthermore, even indexed periodtwo points are saddles and odd indexed periodtwo points are repellers.
Proof The proof is similar as the proof of Corollary 2 and it will be omitted. □
Based on a series of numerical simulations, we propose the following conjecture.
Conjecture 1 System (1) has at most one minimal periodtwo solution.